Section 4.2 The Mean Value Theorem
Theorem 4.2.1. Rolle's Theorem.
Suppose \(y=f(x)\) is a continuous function on \([a,b]\) and differentiable at every point of \((a,b)\text{.}\) If \(f(a)=f(b)\text{,}\) then there is at least one point \(c\) in \((a,b)\) at which \(f'(c)=0\text{.}\)
Proof.
Since \(f\) is continuous on \([a,b]\text{,}\) the Extreme Value Theorem guarantees that \(f\) has both an absolute maximum and an absolute minimum. These extrema can occur only (i) at endpoints of the interval; (ii) at interior points with derivative 0; or (iii) at interior points where the derivative does not exist.
The assumption that \(f\) is differentiable everywhere on \((a,b)\) excludes possibility (iii).
If either extremum occurs on the interior, then by the First Derivative Theorem, the derivative is 0 at that point, which serves as the \(c\) we're looking for.
If both the maximum and minimum occur at the endpoints, the since \(f(a)=f(b)\text{,}\) the function must be constant on \([a,b]\) making the derivative 0 on the entire interval.
Theorem 4.2.2. Mean Value Theorem.
Suppose \(y=f(x)\) is continuous on \([a,b]\) and differentiable on \((a,b)\text{.}\) Then there is at least one \(c\) in \((a,b)\) such that
Proof.
Define the function \(g(x)\) to be the line between the points \((a,f(a))\) and \((b,f(b))\text{.}\) Point-slope form of the equation of a line tells us that
Now define \(h(x)=f(x)-g(x)\text{.}\) Note that \(h\) is continuous on \([a,b]\) and differentiable on \((a,b)\) since both \(f\) and \(g\) are. Also, since \(h(a)=h(b)=0\text{,}\) Rolle's theorem guarantees the existence of a \(c\) in \((a,b)\) such that \(h'(c)=0\text{.}\) But \(h'(x)=f'(x)-g'(x)\) so for this \(c\text{,}\) \(f'(c)=g'(c)\text{.}\) And since \(g\) is a line, its derivative is its slope and we have that \(f'(c)=\frac{f(b)-f(a)}{b-a}\text{.}\)